jmc

The expansion of $(x+m{)}^2-63$ is $x^2+2mx+m^2-63$, which has a constant term of $m^2-63$. This constant term must be equal to the constant term of the original quadratic, so $m^2-63=1$, which yields the possibilities $m=8$ and $m=-8$.

If $m=8$, then $(x+m{)}^2-63=x^2+16x+1$. If $m=-8$, then $(x+m{)}^2-63=x^2-16x+1$. Of these two possibilities, only the first conforms to our information that $b$ was a positive number. So, the original quadratic was $x^2+16x+1$, giving $b=\boxed{16}$.